Server Side Programming Articles - Page 2150 of 2650

Suppose one array A is given. We have to find a number of surpasser of each element in that array. The surpassers are greater elements which are present at the right side of the array of the current element. Suppose A = {2, 7, 5, 3, 0, 8, 1}, the surpassers are {4, 1, 1, 1, 2, 0, 0}, so 2 has 4 numbers at right side, which are greater than 4, and the same rule for others. The solution is very simple, two nested loops will be there, for each element, it will count surpassers, then store them into ... Read More

Here we will see a program, that can find nth term of the Dragon Curve sequence. The Dragon curve sequence is an infinite binary sequence. It starts with 1, and in each step, it alternatively adds 1s and 0s before and after each element of the previous term, to form the next term.Term 1 : 1Term 2 : 110Term 3 : 1101100Term 4 : 110110011100100We will start with 1, then add 1 and 0, alternatively after each element of the preceding term. When the new term obtained becomes the current term, then repeat the steps from 1 to n to ... Read More

Here we will see how to find the LCM of Rational numbers. We have a list of rational numbers. Suppose the list is like {2/7, 3/14, 5/3}, then the LCM will be 30/1.To solve this problem, we have to calculate LCM of all numerators, then gcd of all denominators, then the LCM of rational numbers, will be like −$$LCM =\frac{LCM\:of\:all\:𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟𝑠}{GCD\:of\:all\:𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟𝑠}$$Example Live Demo#include #include #include using namespace std; int LCM(int a, int b) {    return (a * b) / (__gcd(a, b)); } int numeratorLCM(vector vect) {    int result = vect[0].first;    for (int i = 1; i ... Read More

Suppose there are three variables N, R and P. The N and R are used to get the NCR and P is a prime. We have to find whether NCR is divisible by P. Suppose we have some numbers N = 7, R = 2 and P = 3, then 7C2 = 21, this is divisible by 3, so the output will be true.We know that NCR = N! / (R! * (N – R)! ). We will use Legendre Formula to largest power of P, which divides any N!, R! and (N – R)! in order to NCR to ... Read More

Suppose, one circle is given (the center coordinate and radius), another point is also given. We have to find whether the point is inside the circle or not. To solve it, we have to find the distance of the given point from circle center. If that distance is less or equal to the radius, then that is inside the circle, otherwise not.Example Live Demo#include #include using namespace std; bool isInsideCircle(int cx, int cy, int r, int x, int y) {    int dist = (x - cx) * (x - cx) + (y - cy) * (y - cy);    if ( dist

As we know, the HCF or GCD can be calculated easily using the Euclidean Algorithm. But here we will see how to generate GCD or HCF without using the Euclidean Algorithm, or any recursive algorithm. Suppose two numbers are present as 16 and 24. The GCD of these two is 8.Here the approach is simple. If the greater number of these two is divisible by the smaller one, then that is the HCF, otherwise starting from (smaller / 2) to 1, if the current element divides both the number, then that is the HCF.Example Live Demo#include using namespace std; int ... Read More

Suppose we have an array A, with N elements. We have to find the GCD of factorials of all elements of the array. Suppose the elements are {3, 4, 8, 6}, then the GCD of factorials is 6. Here we will see the trick. As the GCD of two numbers, is the greatest number, which divides both of the numbers, then GCD of factorial of two numbers is the value of factorial of the smallest number itself. So gcd of 3! and 5! is 3! = 6.Example Live Demo#include using namespace std; long fact(int n){    if(n arr[i])   ... Read More

Suppose we have three integers A, B, and C. We have to find one minimum integer X, such that X mod C = 0, and X is not in the range [A, B]. If the values of A, B and C are 5, 10 and 4 respectively, then the value of X will be 4. Let us see the steps to get the solution −Steps −If C is not in the range [A, B], then return C as a resultOtherwise get the first multiple of C, which is greater than B, then return that valueExample Live Demo#include using namespace std; ... Read More

Suppose we have an array arr and another value k. We have to find a minimum number of operations to make the GCD of the array equal to the multiple of k. In this case, the operation is increasing or decreasing the value. Suppose the array is like {4, 5, 6}, and k is 5. We can increase 4 by 1, and decrease 6 by 1, so it becomes 5. Here a number of operations is 2.We have to follow these steps to get the result −Steps −for all elements e in the array, follow steps 2 and 3if e ... Read More

Suppose we have two strings of equal length, we have to find a minimum number of alterations required to make two strings anagram, without deleting any character. The Anagram is two strings that have the same set of characters. Suppose two strings are “HELLO”, and “WORLD” here number of required changes is 3, as three characters are different in this case.The idea is simple, we have to find the frequency of each character in the first string, then go through the second string, if characters in the second string are present, in the frequency array, then decrease the frequency value. ... Read More